Factoring into the Product of Two Trinomials

Date: 02/22/2003 at 00:27:36
From: Mark
Subject: Factoring into the product of two trinomials
Write as a product of two trinomials with integer coefficients:
6x^2-xy+23x-2y^2-6y+20
I tried many formulas to help me decode this, but none of them
seemed to apply to this one. Please show this to me step by step,
because I'm not sure if I'm starting correctly.
Thanks for your help.

Date: 02/22/2003 at 14:05:01
From: Doctor Greenie
Subject: Re: Factoring into the product of two trinomials
Hi, Mark -
Here is how I do it....
Let's start by supposing we have found the factorization we want:
6x^2-xy+23x-2y^2-6y+20 = (Ax+By+C)(Dx+Ey+F)
And let's perform the multiplication on the right:
(Ax+By+C)(Dx+Ey+F) = (AD)x^2+(AE+BD)xy+(BE)y^2+(AF+CD)x+(BF+CE)y+(CF)
Now we equate the coefficients of like terms between this product and
the original expression:
AD = 6
AE+BD = -1
BE = -2
AF+CD = 23
BF+CE = -6
CF = 20
Since AD=6 is positive, A and D have the same sign; we can assume they
are both positive. So we have the following possibilities for the
coefficients A and D, since the coefficients are integers:
A D
--------
1 6
2 3
3 2
6 1
And with BE = -2 and again the coefficients being integers, we have
the following possibilities for B and E:
B E
--------
-2 1
-1 2
1 -2
2 -1
Combining each of the possibilities for A and D with each of the
possibilities for B and E, and looking for a combination for which
AE+BD = -1
we find the only combination that works is (A,D) = (2,3) and (B,E) =
(1,-2).
Using these values of A, B, D, and E in the two equations
AF+CD = 23
BF+CE = -6
and solving the resulting pair of linear equations simultaneously, we
find C = 5 and F = 4; and these values satisfy the final equation
CF = 20
Checking our result, we have
(Ax+By+C)(Dx+Ey+F)
= (2x+y+5)(3x-2y+4)
= 6x^2-4xy+8x+3xy-2y^2+4y+15x-10y+20
= 6x^2-xy-2y^2+23x-6y+20
So our calculations were correct.
I hope this helps. Please write back if you have any further questions
about any of this.
- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/